Chem E 260HW #9

C&B: 9.91, 9.94, 9.102E , 9.103 and 10.15E, 10.17, 10.23 ,10.34

9-91 A simple Brayton cycle with air as the working fluid has a pressure ratio of 8. The air temperature at the turbine exit, the net work output, and the thermal efficiency are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats.

Properties The properties of air are given in Table A-17.

Analysis (a) Noting that process 1-2s is isentropic,

Th

Pr1

1

11 5546

= ⏐ →⏐==

310 K310.24 kJ /kg.

( )( )

( )

( )( )( )

kJ/kg 89.167

19.69292.123082.092.1230

K 3.680 and kJ/kg 692.1990.252.2078

1

2.207

kJ/kg 1230.92K 1160

kJ/kg 646.775.0

24.31058.56224.310

K 557.25 and kJ/kg 562.5844.125546.18

433443

43

443

4

33

1212

12

12

221

2

34

3

12

=

−−=

−−=⏐ →⏐−

−=

==⏐ →⏐=⎟⎠

⎞⎜⎝

⎛==

=

=⏐ →⏐=

=−

+=

−+=⏐ →⏐

−

−=

==⏐ →⏐===

sTs

T

ssrr

r

C

ssC

ssrr

hhhhhh

hh

ThPP

PP

P

hT

hhhh

hh

hh

ThPP

PP

ηη

ηη

Thus,

T4 = 770.1 K

(b)

kJ/kg 105.3=−=−=

=−=−=

=−=−=

92.4782.584

/kJ kg 478.9224.31016.789

/kJ kg 584.27.64692.1230

outinout,net

14out

23in

qqwhhqhhq

s

T

1

2s

4s

3qin

qout

1160 K

310 K

2

4

(c) 18.0%===/kJ kg 584.2/kJ kg 105.3

in

out,netth q

wη

9-94 A simple ideal Brayton cycle with air as the working fluid operates between the specified temperature and pressure limits. The net work and the thermal efficiency are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.

Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).

Analysis Using the isentropic relations for an ideal gas,

K .1706kPa 100

kPa 2000K) 300(

0.4/1.4/)1(

1

212 =⎟

⎠

⎞⎜⎝

⎛=⎟⎟⎠

⎞⎜⎜⎝

⎛=

− kk

PP

TT

Similarly,

K 9.424kPa 2000

kPa 100K) 1000(

0.4/1.4/)1(

3

434 =⎟

⎠

⎞⎜⎝

⎛=⎟⎟⎠

⎞⎜⎜⎝

⎛=

− kk

PP

TT

Applying the first law to the constant-pressure heat addition process 2-3 produces

kJ/kg 4.295K)1.7061000)(KkJ/kg 1.005()( 2323in =−⋅=−=−= TTchhq p

Similarly,kJ/kg 5.125K)3009.424)(KkJ/kg 1.005()( 1414out =−⋅=−=−= TTchhq p

The net work production is thenkJ/kg 169.9=−=−= 5.1254.295outinnet qqw

and the thermal efficiency of this cycle is

0.575===kJ/kg 295.4

kJ/kg 169.9

in

netth q

wη

9-102E A simple ideal Brayton cycle with argon as the working fluid operates between the specified temperature and pressure limits. The rate of heat addition, the power produced, and the thermal efficiency are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Argon is an ideal gas with constant specific heats.

Properties The properties of argon at room temperature are R = 0.2686 psiaft3/lbm·R (Table A-1E), cp = 0.1253 Btu/lbm·R and k = 1.667 (Table A-2Ea).

Analysis At the compressor inlet,

/lbmft 670.9psia 15

R) 540)(R/lbmftpsia 2686.0( 33

1

11 =

⋅⋅==

PRT

v

lbm/s 05.62/lbmft 670.9

ft/s) )(200ft 3(3

2

1

11 ===v

VAm&

s

T

1

2

4

3qin

qout

1000 K

300 K

s

T

1

2

4

3qin

qout

1660 R

540 R

According to the isentropic process expressions for an ideal gas,

R 1357psia 15

psia 150R) 540(

70.667/1.66/)1(

1

212 =⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

− kk

PP

TT

R 7.660psia 150

psia 15R) 1660(

70.667/1.66/)1(

3

434 =⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

− kk

PP

TT

Applying the first law to the constant-pressure heat addition process 2-3 gives

Btu/s 2356=−⋅=−= R)13571660)(RBtu/lbm 1253.0(lbm/s) 05.62()( 23in TTcmQ p&&

The net power output is

Btu/s 1417=−+−⋅=

−+−=

R)13575407.6601660)(RBtu/lbm 1253.0(lbm/s) 05.62(

)( 2143net TTTTcmW p&&

The thermal efficiency of this cycle is then

0.601===Btu/s 2356

Btu/s 1417

in

netth

Q

W&

&η

9-103 An aircraft engine operates as a simple ideal Brayton cycle with air as the working fluid. The pressure ratio and the rate of heat input are given. The net power and the thermal efficiency are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.

Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).

Analysis For the isentropic compression process,

K .1527K)(10) 273( 0.4/1.4/)1(12 === − kkprTT

The heat addition is

kJ/kg 500kg/s 1

kW 500inin ===

mQ

q&

&

Applying the first law to the heat addition process,

K 1025KkJ/kg 1.005

kJ/kg 500K 1.527

)(

in23

23in

=⋅

+=+=

−=

p

p

cq

TT

TTcq

The temperature at the exit of the turbine is

K 9.53010

1K) 1025(

10.4/1.4/)1(

34 =⎟⎠

⎞⎜⎝

⎛=⎟⎟⎠

⎞⎜⎜⎝

⎛=

− kk

prTT

Applying the first law to the adiabatic turbine and the compressor producekJ/kg 6.496K)9.5301025)(KkJ/kg 1.005()( 43T =−⋅=−= TTcw p

s

T

1

2

4

3qin

qout273 K

kJ/kg 4.255K)2731.527)(KkJ/kg 1.005()( 12C =−⋅=−= TTcw p

The net power produced by the engine is then

kW 241.2=−=−= kJ/kg)4.2556kg/s)(496. 1()( CTnet wwmW &&

Finally the thermal efficiency is

0.482===kW 500

kW 241.2

in

netth

Q

W&

&η

10-15E A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The rates of heat addition and rejection, and the thermal efficiency of the cycle are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis From the steam tables (Tables A-4E, A-5E, and A-6E),

Btu/lbm 52.13950.102.138Btu/lbm 50.1

ftpsia 5.404

Btu 1 psia)6500)(/lbmft 01645.0(

)(

/lbmft 01645.0

Btu/lbm 02.138

inp,12

33

121inp,

3psia 6 @1

psia 6 @1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

f

f

v

vv

Btu/lbm 4.1120)88.995)(9864.0(02.138

9864.058155.1

24739.08075.1

psia 6

RBtu/lbm 8075.1

Btu/lbm 0.1630

F1200

psia 500

44

44

34

4

3

3

3

3

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgf

fg

f

hxhhs

ssx

ssP

sh

TP

Knowing the power output from the turbine the mass flow rate of steam in the cycle is determined from

lbm/s 9300.0kJ 1

Btu 0.94782

/lbm1120.4)Btu(1630.0

kJ/s 500)(

43

outT,43outT, =⎟

⎠

⎞⎜⎝

⎛−

=−

=⏐ →⏐−=hh

WmhhmW

&&&&

The rates of heat addition and rejection are

Btu/s 913.6

Btu/s 1386

=−=−=

=−=−=

Btu/lbm)02.1380.4lbm/s)(112 9300.0()(

Btu/lbm)52.1390.0lbm/s)(163 9300.0()(

14out

23in

hhmQ

hhmQ

&&

&&

and the thermal efficiency of the cycle is

0.341=−=−=1386

6.91311

in

outth

Q

Q&

&η

10-17 A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The power produced by the turbine and consumed by the pump are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

qin

qout

6 psia1

3

2

4

500 psia

s

T

Analysis From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg 47.25505.442.251kJ/kg 05.4 mkPa 1

kJ 1 kPa)204000)(/kgm 001017.0(

)(

/kgm 001017.0

kJ/kg 42.251

inp,12

33

121inp,

3kPa 20 @1

kPa 20 @1

=+=+==

⎟⎠

⎞⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

f

f

v

vv

kJ/kg 7.2513)5.2357)(9596.0(42.251

9596.00752.7

8320.06214.7

kPa 20

KkJ/kg 6214.7

kJ/kg 3.3906

C700

kPa 4000

44

44

34

4

3

3

3

3

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgf

fg

f

hxhhs

ssx

ssP

sh

TP

The power produced by the turbine and consumed by the pump are

kW 203

kW 69,630

===

=−=−=

kJ/kg) kg/s)(4.05 50(

kJ/kg)7.2513.3kg/s)(3906 50()(

inP,inP,

43outT,

wmW

hhmW

&&

&&

10-23 A 300-MW coal-fired steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits. The overall plant efficiency and the required rate of the coal supply are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

( )( )( )

kJ/kg .0327707.596.271kJ/kg 5.07

mkPa 1

kJ 1kPa 255000/kgm 00102.0

/kgm 000102.0

kJ/kg 96.271

in,12

33

121in,

3kPa 25 @1

kPa 25 @1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

p

p

f

f

whh

PPw

hh

v

vv

( )( ) kJ/kg 2.22765.23458545.096.271

8545.09370.6

8932.08210.6kPa 25

KkJ/kg 8210.6

kJ/kg 2.3317

C450

MPa 5

44

44

34

4

3

3

3

3

=+=+=

=−

=−

=⎭⎬⎫

=

=

⋅=

=

⎭⎬⎫

°=

=

fgf

fg

f

hxhh

s

ssx

ss

P

s

h

T

P

The thermal efficiency is determined from

kJ/kg 2.200496.2712.2276kJ/kg 2.304003.2772.3317

14out

23in

=−=−==−=−=

hhqhhq

and

Thus,( )( )( ) 24.5%==××=

=−=−=

96.075.03407.0

3407.02.3040

2.200411

gencombthoverall

in

outth

ηηηη

ηq

q

(b) Then the required rate of coal supply becomes

and

qin

qout

20 kPa1

3

2

4

4 MPa

s

T

Qi

n

Qou

t

25 kPa1

3

2

4

5 MPa

s

T

·

·

tons/h 150.3==⎟⎟⎠

⎞⎜⎜⎝

⎛==

===

/tons s 04174.0kg 1000

ton1/kJ kg 29,300/kJ s 1,222,992

/kJ s 992,222,12453.0

/kJ s 300,000

coal

incoal

overall

netin

CQm

WQ

&&

&&

η

10-34 A steam power plant that operates on the ideal reheat Rankine cycle is considered. The turbine work output and the thermal efficiency of the cycle are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis From the steam tables (Tables A-4, A-5, and A-6),

( )( )( )

kJ/kg .5425912.842.251

kJ/kg 8.12mkPa 1

kJ 1kPa 208000/kgm 001017.0

/kgm 700101.0

kJ/kg 42.251

in,12

33

121in,

3kPa 20 @1

kPa 20 @1

=+=+=

=⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

p

p

f

f

whh

PPw

hh

v

vv

( )( ) kJ/kg 2.23855.23579051.042.251

9051.00752.7

8320.02359.7kPa 20

KkJ/kg 2359.7

kJ/kg 2.3457

C500

MPa 3

kJ/kg 1.3105MPa 3

KkJ/kg 7266.6

kJ/kg 5.3399

C500

MPa 8

66

66

56

6

5

5

5

5

434

4

3

3

3

3

=+=+=

=−

=−

=

⎭⎬⎫

=

=

⋅=

=

⎭⎬⎫

°=

=

=⎭⎬⎫

=

=

⋅=

=

⎭⎬⎫

°=

=

fgf

fg

f

hxhh

s

ssx

ss

P

s

h

T

P

hss

P

s

h

T

P

The turbine work output and the thermal efficiency are determined from

and( ) ( )

( ) ( ) kJ/kg 0.34921.31052.345754.2595.3399

2.23852.34571.31055.3399

4523in

6543outT,

=−+−=−+−=

=−+−=−+−=

hhhhq

hhhhw kJ/kg 1366.4

Thus,38.9%===

=−=−=

kJ/kg 3492.5

kJ/kg 1358.3

kJ/kg 3.135812.84.1366

in

netth

in,,net

q

w

www poutT

η

1

5

2

6s

T

3

48 MPa

20 kPa